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NNADVOKAT [17]
3 years ago
9

Graph f(x) = |x - 6| - 4

Mathematics
1 answer:
9966 [12]3 years ago
5 0

Answer:

-6 and -4

Step-by-step explanation:

you will go to the box where y is negative and x is negative then 6 will be on the x axis and 4 will be on the y axis

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For part A and B, the options are positive and negative.
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Answer:

Coordinates: (2, -1)

Part A

X coordinate is positive

Y coordinate is negative

Part B

Coordinates after Y-axis reflection: (-2, -1)

X coordiate is negative

Y coordinate is negative

Good luck!

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3 years ago
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Sibal started with $500 in a bank account that does not earn interest. In the middle of every month, she withdraws 15 of the acc
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After every month's withdrawals, 4/5 of the original amount at the start of the month will remain. The amount at the start of every month will change. Thus:
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Name the different types of proofs you saw in this lesson. Give a description for each.
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3 years ago
Solve: 2^x-2^(x-2)=6
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X=3! Hope this helps (:
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3 years ago
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The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. Sh
yulyashka [42]

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 25235

For the alternative hypothesis,

µ > 25235

This is a right tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 27524

µ = population mean = 25235

s = samples standard deviation = 6000

t = (27524 - 25235)/(6000/√100) = 3.815

We would determine the p value using the t test calculator. It becomes

p = 0.000119

Since alpha, 0.05 > than the p value, 0.000119, then we would reject the null hypothesis. There is sufficient evidence to support the claim that student-loan debt is higher than $25,235 in her area.

6 0
3 years ago
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